Are positive definite matrices necessarily diagonalizable and when does the famous eigenvalue criterion apply?

I mean in $\mathbb{C}$ positive definite matrices seem to be self-adjoint. For matrices over real vector spaces this seems to be wrong, but is it still true that they are diagonalizable?

Then everyone knows a result similar to this: When a matrix has only positive eigenvalues then it is positive definite. Is this result always true, and do we also have the converse in general?

Solution 1:

An extension of the definition of "positive definite" to non-symmetric (real) matrices is to require $v^T M v \gt 0$ for all suitable nonzero vectors $v$. However see the discussion on this older Question about the lack of a conventionally adopted definition.

For real matrices this allows for non-self-adjoint (non-symmetric) examples. However the analogous "extension" for complex matrices, $v^* M v \gt 0$ for all nonzero vectors, turns out to imply $M$ self-adjoint (Hermitian), so it gives "nothing new".

As far as diagonalizability goes, neither requiring all eigenvalues positive nor requiring positive definiteness in the sense given above is sufficient to imply similarity to a diagonal matrix. Of course similar to a diagonal matrix would be implied by a full set of distinct eigenvalues (as each eigenvalue has at least one eigenvector), and it is equivalent to a basis of eigenvectors (since the matrix representation wrt to such basis is diagonal).

Here's an example of a nonsymmetric real matrix that (a) has only the positive eigenvalue 1 and (b) satisfies the "coercivity" condition $v^T M v \gt 0$ for all nonzero (real) vectors but (c) cannot be diagonalized:

$$ M = \begin{pmatrix} 1 & 0.1 \\ 0 & 1 \end{pmatrix} $$

Note that if $v^T = (x \; y)$, then $v^T M v = x^2 + 0.1xy + y^2$, which is easily shown to be a positive definite quadratic form, and thus the second (stronger) notion of a positive definite (but nonsymmetric) real matrix holds.